Simplify; express your answer in exponential form. Assume $p\neq 0, q\neq 0$. $\dfrac{{(p^{5})^{-3}}}{{(p^{-2}q^{5})^{5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${p^{5}}$ to the exponent ${-3}$ . Now ${5 \times -3 = -15}$ , so ${(p^{5})^{-3} = p^{-15}}$ In the denominator, we can use the distributive property of exponents. ${(p^{-2}q^{5})^{5} = (p^{-2})^{5}(q^{5})^{5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(p^{5})^{-3}}}{{(p^{-2}q^{5})^{5}}} = \dfrac{{p^{-15}}}{{p^{-10}q^{25}}}$ Break up the equation by variable and simplify. $\dfrac{{p^{-15}}}{{p^{-10}q^{25}}} = \dfrac{{p^{-15}}}{{p^{-10}}} \cdot \dfrac{{1}}{{q^{25}}} = p^{{-15} - {(-10)}} \cdot q^{- {25}} = p^{-5}q^{-25}$.